Question

When a quantity of electricity equal to require to liberate 2.24 L of hydrogen at stp from 0.1M aqueous H2SO4 is passed (at mass of Cu=63.5) then the mass of cu that Will be deposited at cathod in electrolysis of 0.2M solutions of cuso4 will be

Answer 1

Answer: The mass of Cu that will be deposited at the cathode in electrolysis is 6.35 g.

Explanation:

Step 1:

The volume of Hydrogen = 2.24 L

But, at STP the volume of 1 mole of H₂ = 22.4 litres

∴ The no. of moles of hydrogen deposited from H₂SO₄ = 2.24/22.4 = 0.1 moles

Step 2:

Molar mass of H₂ = 2 g/mol …. (i)

∴ Mass of 0.1 moles of H₂ = 2 * 0.1 = 0.2 g …. (ii)

Step 3:

Mass of Cu = 63.5 g .....  [given] …. (iii)

Now, according to Faraday's second law of electrolysis, we know that, when the same quantity of electricity is passed through several electrolytes, the mass of the substances that are deposited are proportional to their respective equivalent weight.

Therefore,

[Weight of Cu] / [weight of H₂] = [eq. mass of Cu] / [eq. mass of H₂]

Substituting the values from (i), (ii) & (iii), we get

⇒ Weight of Cu = [63.5/2] * 0.2 = 6.35 g